In the following redox reaction P4(s) + 3OH-(aq)+3H2O(l)→PH3(g) + 3H2PO2-(aq), the oxidation state of phosphorus changes from

Phosphorus changes 0 to -3 and 0 to +1.This is a disproportionation: elemental phosphorus, P4, is in the 0 state.In PH3: \(x + 3(+1) = 0\), so P is -3.In H2PO2-: \(2(+1) + x + 2(-2) = -1\), so P is +1.So part of the phosphorus is reduced to -3 and part oxidised to +1. The option 0 to +3 would need phosphite, H2PO3-, and 0 to +5 would need phosphate; neither is formed here.

Using stock notation, match the oxidation number against metal in its compound.CompoundOxidation Number of the metalMnO(a) IVMnO2(b) IIFe2O3(c) ITl2O(d) III

The matching is (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c), taking the compounds in the order listed.With oxygen at -2: MnO has Mn(II), so manganese(II) oxide (b); MnO2 has Mn(IV), manganese(IV) oxide (a); Fe2O3 has 2x = +6, so Fe(III), iron(III) oxide (d); Tl2O has 2x = +2, so Tl(I), thallium(I) oxide (c).Stock notation writes this oxidation number in Roman numerals after the metal's name. The other options swap Fe2O3 and MnO2, or give Tl2O a higher state, which the formula does not allow.

Redox Reactions Questions and Answers

Q: In the following reaction, the change in oxidation state of Magnesium is\(\mathrm{3Mg_{(s)} + N_{2(g)} \xrightarrow{\Delta} Mg_{3}N_{2(s)}}\)

Magnesium changes from 0 to +2.Free magnesium metal, \(\mathrm{Mg_{(s)}}\), is an element in its standard state, so its oxidation state is 0. In magnesium nitride, \(\mathrm{Mg_{3}N_{2}}\), nitrogen is the more electronegative element and takes -3:\[ \begin{aligned} 3x + 2(-3) &= 0 \\ x &= +2 \end{aligned} \]So each magnesium atom loses two electrons and is oxidised, while nitrogen goes from 0 to -3 and is reduced. +3, +4 and +6 are not oxidation states of magnesium, a group 2 metal that only ever shows +2 in compounds, and a negative value would mean magnesium had gained electrons.

Q: Which of the following statement is INCORRECT for the concept of oxidation?

The INCORRECT statement is Decreases in number of positive charges.Oxidation is the loss of electrons. Losing an electron makes a species more positive: Fe2+ → Fe3+ + e- increases the positive charge, and Cl- → Cl decreases the negative charge. A decrease in positive charge is the opposite, a reduction.The other statements describe oxidation in its classical and electronic senses: addition of oxygen, removal of hydrogen, removal of an electron, and the decrease in negative charge that follows from losing electrons.

Q: Which of the following is INCORRECT for the concept of reduction?

The incorrect statement is Removal of an electron.Reduction is the gain of electrons; removal (loss) of an electron is oxidation. The other four are all valid descriptions of reduction:removal of oxygen, as in CuO + H2 giving Cu,addition of hydrogen, as in ethene giving ethane,addition of an electron, as in Fe3+ + e- giving Fe2+,decrease in oxidation number, which is what electron gain does to the atom.

Q: In which of the following compound oxygen is in -1 oxidation state?

Oxygen is in the -1 state in hydrogen peroxide.In H2O2 the two oxygen atoms are bonded to each other (a peroxide link, -O-O-). Each hydrogen is +1, so the two oxygens share a total of -2, which is -1 each.In water, manganese dioxide, carbon dioxide and potassium nitrate oxygen has its usual oxidation state of -2. The exceptions to -2 are peroxides (-1), superoxides (-1/2) and compounds with fluorine, such as OF2 (+2).

Q: The oxidation numbers of carbon and hydrogen in methane are respectively

The oxidation numbers are -4 and +1.In CH4, hydrogen bonded to a non-metal has oxidation number +1. The molecule is neutral, so the oxidation numbers must add to zero:\[ \begin{aligned} x + 4(+1) &= 0 \\ x &= -4 \end{aligned} \]Carbon is more electronegative than hydrogen, which is why it carries the negative value. Any option giving carbon +4 or hydrogen 0 or -1 breaks this rule; hydrogen is -1 only in metal hydrides such as NaH.

Q: The oxidation number of oxygen in O2, O2F2 and RbO2 are respectively

The oxidation numbers are 0, +1, -1/2.In \(\mathrm{O_2}\) oxygen is an element, so 0. In \(\mathrm{O_2F_2}\) fluorine, the most electronegative element, is always -1, so the two oxygens share +2: each is +1. In \(\mathrm{RbO_2}\), a superoxide, Rb is +1, so the two oxygens share -1: each is -1/2.\[ \begin{aligned} \mathrm{O_2F_2}: 2x + 2(-1) &= 0 \Rightarrow x = +1 \\ \mathrm{RbO_2}: +1 + 2x &= 0 \Rightarrow x = -\tfrac{1}{2} \end{aligned} \]The option with +2 uses \(\mathrm{OF_2}\) instead of \(\mathrm{O_2F_2}\); -1 is the value in peroxides, not superoxides.

Q: Chlorine has oxidation number +7 in

Chlorine is +7 in HClO4.In perchloric acid, hydrogen is +1 and each of the four oxygens is -2: 1 + x + 4(-2) = 0, so x = +7, chlorine's highest oxidation state.In NaCl, KCl, HCl and SiCl4 chlorine is bonded to a less electronegative element and has the oxidation number -1. In general chlorine's positive states appear only with oxygen or fluorine: +1 in HClO, +3 in HClO2, +5 in HClO3 and +7 in HClO4.

In the following reaction, the change in oxidation state of Magnesium is

\(\mathrm{3Mg_{(s)} + N_{2(g)} \xrightarrow{\Delta} Mg_{3}N_{2(s)}}\)

A.0 to +2
B.0 to +3
C.0 to +4
D.0 to +6
E.0 to −2

Exam Year(s)

KEAM - 2025,

Answer: A

Magnesium changes from 0 to +2.

Free magnesium metal, \(\mathrm{Mg_{(s)}}\), is an element in its standard state, so its oxidation state is 0. In magnesium nitride, \(\mathrm{Mg_{3}N_{2}}\), nitrogen is the more electronegative element and takes -3:

\[ \begin{aligned} 3x + 2(-3) &= 0 \\ x &= +2 \end{aligned} \]

So each magnesium atom loses two electrons and is oxidised, while nitrogen goes from 0 to -3 and is reduced. +3, +4 and +6 are not oxidation states of magnesium, a group 2 metal that only ever shows +2 in compounds, and a negative value would mean magnesium had gained electrons.

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Which of the following statement is INCORRECT for the concept of oxidation?

A.Addition of oxygen
B.Removal of hydrogen
C.Decreases in number of positive charges
D.Decreases in number of negative charges
E.Removal of an electron

Exam Year(s)

KEAM - 2025,

Answer: C

The INCORRECT statement is Decreases in number of positive charges.

Oxidation is the loss of electrons. Losing an electron makes a species more positive: Fe²⁺ → Fe³⁺ + e^- increases the positive charge, and Cl^- → Cl decreases the negative charge. A decrease in positive charge is the opposite, a reduction.

The other statements describe oxidation in its classical and electronic senses: addition of oxygen, removal of hydrogen, removal of an electron, and the decrease in negative charge that follows from losing electrons.

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Which of the following is INCORRECT for the concept of reduction?

A.Removal of oxygen
B.Addition of hydrogen
C.Addition of electron
D.Decrease in oxidation number
E.Removal of an electron

Exam Year(s)

KEAM - 2025,

Answer: E

The incorrect statement is Removal of an electron.

Reduction is the gain of electrons; removal (loss) of an electron is oxidation. The other four are all valid descriptions of reduction:

removal of oxygen, as in CuO + H₂ giving Cu,
addition of hydrogen, as in ethene giving ethane,
addition of an electron, as in Fe³⁺ + e^- giving Fe²⁺,
decrease in oxidation number, which is what electron gain does to the atom.

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In which one of the following reactions, +2 state in a reactant is reduced to +1 state?

A.\(\mathrm{Mg + S \rightarrow MgS}\)
B.\(\mathrm{2HgCl_{2} + SnCl_{2} \rightarrow Hg_{2}Cl_{2} + SnCl_{4}}\)
C.\(\mathrm{CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O}\)
D.\(\mathrm{2K_{4}[Fe(CN)_{6}] + H_{2}O_{2} \rightarrow 2K_{3}[Fe(CN)_{6}] + 2KOH}\)
E.\(\mathrm{2H_{2}S + O_{2} \rightarrow S + 2H_{2}O}\)

Exam Year(s)

KEAM - 2026,

Answer: B

The reaction is \(\mathrm{2HgCl_{2} + SnCl_{2} \rightarrow Hg_{2}Cl_{2} + SnCl_{4}}\).

In HgCl₂ mercury is +2. In Hg₂Cl₂ (calomel, with an Hg-Hg bond) each mercury is +1. So the +2 state is reduced to +1, while tin is oxidised from +2 to +4.

The others involve different changes: in Mg + S, Mg goes 0 to +2; in the combustion of methane carbon goes -4 to +4; in the ferrocyanide reaction iron goes +2 to +3, an oxidation; and in H₂S + O₂ sulphur goes -2 to 0. None of them reduces +2 to +1.

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In the following redox reaction P_4(s) + 3OH^-_(aq)+3H₂O_(l)→PH_3(g) + 3H₂PO₂^-_(aq), the oxidation state of phosphorus changes from

A.0 to -3 and 0 to -1
B.0 to +1 and 0 to +3
C.0 to -1 and 0 to +5
D.0 to -3 and 0 to +1
E.-3 to 0 and +1 to +3

Exam Year(s)

KEAM - 2026,

Answer: D

Phosphorus changes 0 to -3 and 0 to +1.

This is a disproportionation: elemental phosphorus, P₄, is in the 0 state.

In PH₃: \(x + 3(+1) = 0\), so P is -3.
In H₂PO₂^-: \(2(+1) + x + 2(-2) = -1\), so P is +1.

So part of the phosphorus is reduced to -3 and part oxidised to +1. The option 0 to +3 would need phosphite, H₂PO₃^-, and 0 to +5 would need phosphate; neither is formed here.

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In which of the following compound oxygen is in -1 oxidation state?

A.Water
B.Manganese dioxide
C.Hydrogen peroxide
D.Carbon dioxide
E.Potassium nitrate

Exam Year(s)

KEAM - 2026,

Answer: C

Oxygen is in the -1 state in hydrogen peroxide.

In H2O2 the two oxygen atoms are bonded to each other (a peroxide link, -O-O-). Each hydrogen is +1, so the two oxygens share a total of -2, which is -1 each.

In water, manganese dioxide, carbon dioxide and potassium nitrate oxygen has its usual oxidation state of -2. The exceptions to -2 are peroxides (-1), superoxides (-1/2) and compounds with fluorine, such as OF2 (+2).

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The oxidation numbers of carbon and hydrogen in methane are respectively

A.+4 and +1
B.+4 and 0
C.-4 and +1
D.-4 and -1
E.+4 and -1

Exam Year(s)

KEAM - 2026,

Answer: C

The oxidation numbers are -4 and +1.

In CH₄, hydrogen bonded to a non-metal has oxidation number +1. The molecule is neutral, so the oxidation numbers must add to zero:

\[ \begin{aligned} x + 4(+1) &= 0 \\ x &= -4 \end{aligned} \]

Carbon is more electronegative than hydrogen, which is why it carries the negative value. Any option giving carbon +4 or hydrogen 0 or -1 breaks this rule; hydrogen is -1 only in metal hydrides such as NaH.

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The oxidation number of oxygen in O₂, O₂F₂ and RbO₂ are respectively

A.0, +1, +2
B.0, +2, -1/2
C.0, +1, -1/2
D.0, 0, +1/2
E.0,-1, -1/2

Exam Year(s)

KEAM - 2026,

Answer: C

The oxidation numbers are 0, +1, -1/2.

In \(\mathrm{O_2}\) oxygen is an element, so 0. In \(\mathrm{O_2F_2}\) fluorine, the most electronegative element, is always -1, so the two oxygens share +2: each is +1. In \(\mathrm{RbO_2}\), a superoxide, Rb is +1, so the two oxygens share -1: each is -1/2.

\[ \begin{aligned} \mathrm{O_2F_2}: 2x + 2(-1) &= 0 \Rightarrow x = +1 \\ \mathrm{RbO_2}: +1 + 2x &= 0 \Rightarrow x = -\tfrac{1}{2} \end{aligned} \]

The option with +2 uses \(\mathrm{OF_2}\) instead of \(\mathrm{O_2F_2}\); -1 is the value in peroxides, not superoxides.

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Using stock notation, match the oxidation number against metal in its compound.

Compound	Oxidation Number of the metal
MnO	(a) IV
MnO₂	(b) II
Fe₂O₃	(c) I
Tl₂O	(d) III

A.(i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
B.(i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)
C.(i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)
D.(i)-(d), (ii)-(a), (iii)-(b), (iv)-(c)
E.(i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)

Exam Year(s)

KEAM - 2026,

Answer: A

The matching is (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c), taking the compounds in the order listed.

With oxygen at -2: MnO has Mn(II), so manganese(II) oxide (b); MnO₂ has Mn(IV), manganese(IV) oxide (a); Fe₂O₃ has 2x = +6, so Fe(III), iron(III) oxide (d); Tl₂O has 2x = +2, so Tl(I), thallium(I) oxide (c).

Stock notation writes this oxidation number in Roman numerals after the metal's name. The other options swap Fe₂O₃ and MnO₂, or give Tl₂O a higher state, which the formula does not allow.

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10.

Chlorine has oxidation number +7 in

A.NaCl
B.\(\mathrm{SiCl_{4}}\)
C.HCl
D.\(\mathrm{HClO_{4}}\)
E.KCl

Exam Year(s)

KEAM - 2026,

Answer: D

Chlorine is +7 in HClO₄.

In perchloric acid, hydrogen is +1 and each of the four oxygens is -2: 1 + x + 4(-2) = 0, so x = +7, chlorine's highest oxidation state.

In NaCl, KCl, HCl and SiCl₄ chlorine is bonded to a less electronegative element and has the oxidation number -1. In general chlorine's positive states appear only with oxygen or fluorine: +1 in HClO, +3 in HClO₂, +5 in HClO₃ and +7 in HClO₄.

Redox Reactions Questions and Answers

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